Question

The value of $\sec (2{\cot }^{-1}2+{\cos }^{-1}\frac{3}{5})$ is equal to

• A. $\frac{25}{24}$
• B. $-\frac{24}{7}$
• C. $\frac{25}{7}$
• D. $-\frac{25}{7}$

Answer 1

The correct option is D $-\frac{25}{7}$

Let $\alpha =2{\cot }^{-1}2+{\cos }^{-1}\frac{3}{5}$
We know that ${\cot }^{-1}x={\tan }^{-1}\frac{1}{x}$ for $x>0$
and $2{\tan }^{-1}x={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$ for $|x|<1$
$\therefore \alpha =2{\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{4}{3}$
$={\tan }^{-1}\left(\frac{2\times \frac{1}{2}}{1-{\left(\frac{1}{2}\right)}^2}\right)+{\tan }^{-1}\frac{4}{3}$
$\Rightarrow \alpha =2{\tan }^{-1}\frac{4}{3}$
$\Rightarrow \frac{\alpha }{2}={\tan }^{-1}\frac{4}{3}$
$\Rightarrow \tan \left(\frac{\alpha }{2}\right)=\frac{4}{3}$

Now, $\sec \alpha =\frac{1+{\tan }^2\left(\frac{\alpha }{2}\right)}{1-{\tan }^2\left(\frac{\alpha }{2}\right)}$
$=\frac{1+\left(16/9\right)}{1-\left(16/9\right)}=-\frac{25}{7}$